Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 33

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Question 1
1 / -0

$$(X)$$ on treatment with sodium hydroxide followed by the addition of silver nitrate gives white precipitate at room temperature which is soluble in $${NH}_{4}OH$$. $$(X)$$ can be:

A
chlorobenzene

B
ethyl bromide

C
benzyl chloride

D
vinyl chloride

Solution

Silver nitrate solution can be used to find out which halogen is present in a suspected halogenoalkane. $$NaOH$$ substitutes the $$Cl^-$$ from haloalkane.
In Chlorobenzene and Vinyl chloride, the $$Cl$$ is in conjugation with benzene ring and does not give alcohol.
In Benzylchloride, the $$Cl$$ is not directly linked to benzene and thus react as haloalkane and gives white precipitate of $$AgCl$$ with $$AgNO_3$$ which is soluble in $$NH_4OH$$.
Ethyl bromide gives yellow precipitate of $$AgBr$$ which is insoluble in $$NH_4OH$$ at room temperature.
Question 2
1 / -0

Which of the following halides can yield ethane and also methane in a single step?

A
$${C}_{2}{H}_{5}Br$$

B
$${CH}_{3}I$$

C
$${ \left( { CH }_{ 3 } \right) }_{ 2 }CHBr$$

D
None

Solution

Hence the answer is (b)
Question 3
1 / -0

$${CH}_{3}-CH(Cl)-{C}_{2}{H}_{5} \xrightarrow [ ]{ Alc.\quad KOH } {CH}_{3}-CH=CH-{CH}_{3}$$
The above reaction proceeds via $$E_1cb$$ mechanism?

A
It is second order and bumolecular

B
It is first order and unimolecular

C
It is first order an bimolecular

D
It is second order and unimolecular

Solution

The given reaction (dehydrochlorination of $$2$$-chlorobutane in presence of alcoholic $$KOH$$ to form $$2$$-butene) proceeds via $$E_1cb$$ mechanism.
$$E_1$$ = unimolecular elimination ;cB = conjugate base.
It is second order and unimolecular.
Question 4
1 / -0

Neopentyl chloride on reaction with ethanolic $$KOH$$ is likely to:

A
Neopentyl alcohol

B
Pentylene

C
$$2$$-Methyl-$$2$$-butene

D
Undergo no reaction

Solution

Neopentyl chloride on reaction with ethanolic $$KOH$$ is likely to undergo no reaction as it lacks alpha $$H$$ atom. Due to this elimination is not possible.

Option D is correct.
Question 5
1 / -0

Which of the following reactions is not stereospecific?

A
$${S_N}{2}$$

B
Addition of $${Br}_{2}$$ to ethylene in $$C{Cl}_{4}$$

C
Electrophilic substitution

D
Glycol formation from alkenes with alkaline $$KMn{O}_{4}$$

Solution

a. $${S_N}{2}$$ is stereospecific and stereoselective
b. Addition of $${Br}_{2}$$ is stereospecific i.e anti-addition of two $$Br$$ atoms.
c. Electrophilic substitution is not stereospecific.
d. Hydroxylation is stereopsecific, syn-addition of two $$(-OH)$$ groups.
Question 6
1 / -0

Which of the following represents Westrosol?

A
$$CH{Cl}_{3}$$

B
$${CH}_{2}{Cl}_{2}$$

C
$$CH{Cl}_{2}{CH}_{2}Cl$$

D
$$C{Cl}_{2}=CHCl$$

Solution

When westron vapours are passed over heated $$Ba{Cl}_{2}$$ or lime, westrol (trichloro ethylene) is obtained.
$${Cl}_{2}CH-CH{Cl}_{2}\xrightarrow [ Ba{ Cl }_{ 2 } ]{ } {Cl}_{2}C=CHCl$$ (Westrol)
Question 7
1 / -0

In the reaction of p-chlorotoluene with $$K{NH}_{2}$$ in liquid $${NH}_{3}$$, the major product is:

A
o-toluidine

B
m-toluidine

C
p-toluidine

D
p-chloroaniline

Solution

Hence the answer is (b)
Question 8
1 / -0

The reaction of toluene with chlorine in the presence of ferric chloride gives predominantly:

A
Benzoyl chloride

B
$$m-chlorotoluene$$

C
Benzyl chloride

D
$$o-$$ and $$p-chlorotoluene$$

Solution

In presence of $$FeCl_3$$ and in dark, electrophilic substitution on benzene ring by $$Cl^+$$ on at ortho and para positions occurs to give o- and p-chlorotoluene.
Since the methyl group activates the ring at ortho and para position more than meta position, the electrophilic substitution occurs mostly at these positions.
Again, among these, the para isomer is the major product due to sterric reasons.
Hence the answer is (d)
Question 9
1 / -0

The reaction of $$t$$-butyl bromide with sodium methoxide mainly produces:

A
isobutane

B
isobutylene

C
$$t$$-butyl methyl ether

D
sodium tert-butoxide

Solution

Hence the answer is (b)
Question 10
1 / -0

$$n$$-Propyl bromide on treatment with ethanolic potassium hydroxide produces:

A
propane

B
propene

C
propyne

D
propanol

Solution

$$\underset { n-Propyl\:bromide }{ { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }Br } \xrightarrow [ ]{ Ethanolic\ KOH } \underset { Propene }{ { CH }_{ 3 }CH={ CH }_{ 2 } } +KBr+{ H }_{ 2 }O$$